We observe that /_BCA=10° while /_ADB =20° & simultaneously /_ADC=155° =(Larger /_CDA)/2.Hence would it be correct to say that the circumcircle of Tr. ABC will have point D as its centre and DA(=DB=DC) as its radius? If that be correct then /_BAC = x = /_BDC/2 =30/2=15°.